\begin{answer}
Note it also states that $p(x|\theta) = p(x)$. We have

$$
    \begin{aligned}
        \theta_{\text{MAP}} &= \arg \max_{\theta}p(\theta| x, y)\\
        &= \arg \max_\theta p(x, y|\theta) p(\theta)\\
        &= \arg \max_\theta p(y|x, \theta) p(x|\theta)p(\theta) \\
        &= \arg \max_\theta p(y|x, \theta)p(\theta)
    \end{aligned}
$$

\end{answer}
